4
$\begingroup$

Prove that $C^{\infty}_ {c}(\mathbb{R}^n)$ is dense in $W^{k,p}(U)$ for any open $U\subset \mathbb{R}^n$ with $\partial U\in C^1.$ In which $p\in [1,\infty)$

Note: In Lawrence Evans's PDE text, the case where $U$ is bounded was proved as a Theorem. It thus only remained to prove the case when $U$ is unbounded.

Given $f \in W^{k,p}(U).$ I am thinking about partition $U$ into a sequence of annulli. Let $A_n:= \{x\in \mathbb{R}^n:n-1<|x|<n\},\forall n\in \mathbb{N}$ and let $U_n:=A_n\cap U.$ Then each $U_n$ satisfies the hypothesis of the Theorem for being a bounded set, and thus on each $U_n$ there exists a sequence $(f^{(n)}_j:j\in \mathbb{N}) \subset C^{\infty}_ {c}(U_n)$ which converges to the restriction $f\chi_{U_n}.$

But, how does one guarentee that the "glued-fuction" $\sum_{n=1}^{\infty}f^{(n)}_j\chi_{U_n}$ for each fixed $j$ is in $C^{\infty}_ {c}(U)$?? Is my idea plausible? If so, how to make it rigorous?

Thanks for any feedback!

$\endgroup$
  • 1
    $\begingroup$ The first name is Lawrence, the last name is Evans. And he is usually called "Craig". $\endgroup$ – Hans Engler Dec 2 '14 at 4:01
  • $\begingroup$ I don't think your method will work. Even if each $f_j$ is good, but after you glue it you will end up with a $C^\infty$ function. You are actually repeat the prove of Meyers-Serrin theorem. Note that $\partial U$ is $C^1$ enable us to choose countably many open ball $B_n$, centered at $\partial U$, such that $(B_n)$ is an open cover of $\partial U$ and we choose $phi_n$ be the partition of unity subordinated to $(B_n)$. Then you should define your approximation function in each of balls. Sorry i'm tired tonight. I will write an answer for you tmr if you still need it $\endgroup$ – spatially Dec 3 '14 at 3:19
  • 1
    $\begingroup$ This is false. The closure of $C^\infty_c$ in $W^{k,p}$ is usually denoted by $W^{k,p}_0$. This is a proper subspace of $W^{k,p}$ if the domain has a boundary. (So it is true that $W^{k,p}_0(\mathbb{R}^n)=W^{k,p}(\mathbb{R}^n)$, but not that $W^{k,p}_0(U)=W^{k,p}(U)$ is $U$ is a proper domain). $\endgroup$ – Giuseppe Negro Dec 3 '14 at 15:49
  • 1
    $\begingroup$ I think what he wants is $C_c^{\infty}(R^N)$ but not $C_c^\infty(U)$, as he is trying to extend the density theorem by Evans's book. $\endgroup$ – spatially Dec 3 '14 at 19:21
  • $\begingroup$ Thanks for pointing that out. It should be $C^{\infty}_{c}(\mathbb{R}^n)$ instead of $C^{\infty}_{c}(U).$ I have just corrected it. $\endgroup$ – user31899 Dec 5 '14 at 16:05
2
$\begingroup$

There are two ways we could fix this.

The fastest way is noticing that $U$ is an extension domain and we could extend $u$ to $\bar{u}\in W^{1,p}(R^N)$ such that $\bar{u}=u$ inside $U$.

Next, we could use $(v_n)\subset C_c^{\infty}(R^N)$ to approximate $\bar{u}$ in $W^{1,p}$ by the fact that $W_0^{1,p}(R^N)=W^{1,p}(R^N)$. Then $v_n$ restrict to $U$ will do the job for you.

Note that generally we require $U$ to be bounded to be an extension domain, like the one you read from Evans or H. Brezis, but we do can do extension on unbounded domain, with more complicated approach. You could find one from Leoni's book, Theorem 12.15.

Moreover, you could also follows the idea in the comment I wrote yesterday, which only require $\partial U$ to be continuous. You could also find details in Theorem 10.29 in Leoni's book.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.